# How do you implicitly differentiate 6=ylny-e^x-e^y?

Jan 8, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} / \left(\ln y + 1 - {e}^{y}\right)$

#### Explanation:

Find the derivative of each part.

$\frac{d}{\mathrm{dx}} \left(6\right) = 0$

Use the product rule and chain rule here.

$\frac{d}{\mathrm{dx}} \left(y \ln y\right) = y ' \ln y + y \left(\frac{1}{y}\right) y ' = y ' \ln y + y '$

$\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$

Chain rule:

$\frac{d}{\mathrm{dx}} \left({e}^{y}\right) = y ' {e}^{y}$

Thus, the derivative of the entire implicit equation is

$0 = y ' \ln y + y ' - {e}^{x} - y ' {e}^{y}$

Solve for $y '$.

${e}^{x} = y ' \left(\ln y + 1 - {e}^{y}\right)$

$y ' = {e}^{x} / \left(\ln y + 1 - {e}^{y}\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$